When a mass(m) is attached to a vertical spring(k) and slowly brought to its equilibrium position, what will it's extension(say x) be?
Actually both the answers are correct. If you want to find the extension in spring when the block is in equilibrium then you should write an equation making net force on the block equal to zero. So in equilibrium:Kx=mg => x=mg/k.But if you drop the block when the spring is in its natural length then block will go upto x=2mg/k because when it will pass through its equilibrium stage its acceleration would be zero but its velocity will not be zero hence it will go down further until its velocity becomes zero. Hence by applying work energy theorem:Mgx=(1/2)kx^2 => x=2Mg/k. So if you will let the block come down slowly then elongation of spring will be Mg/k but if you will just drop the block after attaching it to spring then elongation will be 2Mg/k. Hope you understand.
Please help me answer some questions about elephants!?
please answer these questions1.Describe how and elephant moves (fast slow, what structures or features does it have to enable it to move?2.How does an elephant's movement help it survive? Explain3.What types of food does an elephant eat?4.How often does an elephant need to eat?5.Describe an elephants digestive system or organs. 6.Does an elephant expend a lot of energy each day? explain.7.How large is it as an adult?8.What stages does it go through from birth to adulthood9.What is the life span? 10.What is the average life expectancy?11.What attracts it? How does it respond?12.How does it protect itself from enemies who are its enemies?13.How does it find a mate, is there competition is ther courting14.describe birth process15.what is the climate like where it lives 16.what body structures or features does it have to survive in its environment17.what is unique about it that has allowed its species to survive over long periods of time 18.how does it interact w/ 2 other organisms thanx!
How to find the unknown weight of any object using density factor?
im a rigger (work with crains) i know it starts lenth by width by hight miltiplied by 500 then the density factor of .02,.03 ect. or something close to this equation
A boy of mass 40 kg is standing on a weighing machine fixed to the floor of elevator. When the elevator begins to move up with constant acceleration a, the machine shows weight of the boy to be equal to 800 N. What's the value of a?
Based on Newton's second law. "Net force on a body equals product of its mass and acceleration." We need to find the net force I.e the resultant of all forces on the boy.We have been given the following forces1. Gravitational force downward =40g ~400N2. Normal reaction upward on the boy from the weighing machine=weight of the boy shown on the weighing machine=800NNet upward force on the boy=total external upward force-total external downward force=800N-400N=400NNet upward acceleration(again from second law)= 400/40= 10 m/s^2
If a block of mass (m) is kept on a horizontal surface with friction coefficient [math]\mu[/math], then what would be the minimum force needed to move that block on the surface?
This is actually a tricky one since many people will end up gettingminimum force as [math]f = \mu.mg [/math]However Just read the question again , it says minimum force (There is no mention of minimum horizontal force)Now as we know Friction [math]force \sim Normal[/math]i.e. : if we minimize the normal reaction we can minimize the friction at the same time we also need to provide enough Horizontal force to ensure block gets a horizontal velocitySomething like shown in figure !Now the task is to find optimal angle “x” and Magnitude FApplying simple physics : (Assuming Normal reaction from surface is N)Balancing Vertical force[math]Fsinx+N = mg [/math] —>(1)[math]N = mg - Fsinx[/math]Balancing Horizontal force[math]Fcosx = Friction[/math]=> [math]Fcosx = \mu.N[/math]=> [math]Fcosx = \mu.(mg-Fsinx) [/math]=> [math]F(cosx+\mu.sinx) = \mu.mg [/math]=> [math]F = \frac{\mu.mg}{cosx+\mu.sinx} [/math] ==>(2)Now to minimize F(2) we need to maximize denominator of 2[math]max[/math]{[math]cosx+\mu.sinx[/math]}[math] = \sqrt{1+(\mu)^2} [/math] Basic TrigoSo minimum force needed would be[math]F = \frac{\mu.mg}{\sqrt{1+(\mu)^2}}[/math]
If the earth is rotating at a high speed and we jump up, why doesn't the earth move below us at high speed?
Since other answers have correctly mentioned the atmosphere and such, let's take it out of the equation for the sake of argument and see what happens.In terms of the difference in the helicopter's path vs. the ground, let's look at what happens during one whole day as the Earth rotates. Assuming the helicopter can perform an ideal, vertical-acceleration-only hover, then the surface of the Earth and the helicopter have the same tangential speed as the Earth turns, due to conservation of momentum. But, since the helicopter is now a few feet (~1 meter) above the ground, the path the helicopter takes to go around the Earth during one day is now slightly longer as compared to the surface itself. The added circumference of that trip all the way around the Earth for the helicopter turns out be just 2 times Pi times the height above the ground. It takes the Earth just under 24 hours to rotate once on its axis, so the helicopter would have to move at an additional horizontal speed of maybe 18 feet (~5.5 meters) per 24 hours, or something like 0.00014 miles per hour (~0.00022 km/h) to stay directly over the same spot on the ground. This is such a negligible difference in horizontal speed that you don't notice it in any practical system.A real helicopter would deviate horizontally by a speed that is orders of magnitude more than that when it takes off anyway. And, of course, helicopters fly in the atmosphere, which is coupled to the ground. It's also fair to say that wind speed dominates any effects from conservation of momentum, as it's orders of magnitude larger, even at high altitudes. So, real aircraft worry about the difference in wind speed and direction vs. the ground, rather than orbital mechanics.